How To Write An Effective Online Job Posting

How To Write An Effective Online Job Posting

English
17 Pages
Read
Download
Downloading requires you to have access to the YouScribe library
Learn all about the services we offer

Description

  • expression écrite
How To Write An Effective Online Job Posting November, 2010 “An effective job posting saves time and money by driving the best and most qualified candidates to you fast!” by Marc Sampson CEO - BirdDog
  • push applicants into a replicable system
  • tracking of data
  • internal email address like resumes
  • targeted response
  • career path
  • job
  • candidates
  • applicants
  • skills

Subjects

Informations

Published by
Reads 17
Language English
Report a problem

MA441: Algebraic Structures I
Lecture 23
1 December 2003
1Review from Lecture 22:
Definition:
A subgroup H of a group G is called a normal if aH =Ha for all a∈G.
We denote this by HCG.
Theorem 9.1: Normal Subgroup Test
A subgroup H of G is normal in G iff
1xHx H for all x∈G.
2Definition:
For HCG, the set of left (or right) cosets of
H in G is a group called the quotient group
(or factor group) of G by H.
We denote this group by G/H.
Theorem 9.2:
Let HCG. The set G/H is a group under the
operation (aH)(bH)=abH.
3Example 11:
Let G=U(32)={1,3,5,7,...,27,29,31}.
LetH =U (32)={1,17}. (171 (mod 16))16
G/H is abelian of order 16/2=8.
In Chapter 11, we’ll learn about the Funda-
mental Theorem of Abelian Groups. This the-
orem tells us that an abelian group of order 8
must be isomorphic to Z , Z Z , or8 4 2
Z Z Z .2 2 2
Which of these three direct products is isomor-
phic to G/H?
4We will determine the elements of G/H and
their orders.
1H ={1,17}, 3H ={3,19}, 5H ={5,21}
7H ={7,23}, 9H ={9,25}, 11H ={11,27}
13H ={13,29}, 15H ={15,31}.
2We can rule out Z Z Z , because (3H) =2 2 2
9H, so |3H|4.
2 2Now (7H) = (9H) = H, so these are two
distinct elements with order 2. This rules out
Z , which has only one element of order 2.8
Therefore, U(32)/U (32) Z Z , which is16 4 2
isomorphic to U(16).
5Theorem 9.3:
Let G be a group with center Z(G). If G/Z(G)
is cyclic, then G is Abelian.
Proof:
Letg ∈GbesuchthatgZ(G)generatesG/Z(G).
For any a,b∈G, let
i iaZ(G)=(gZ(G)) =g Z(G),
and let
j jbZ(G)=(gZ(G)) =g Z(G).
6i jThen a=g x and b=g y, ∃x,y ∈Z(G).
iSince x,y commute with everything, and g
jcommutes with g , we see that
i j i+j j iab=(g x)(g y)=g xy =(g y)(g x)=ba.
Since the a,b were arbitrary, we have that G is
abelian.
7Theorem 9.4:
For any group G, G/Z(G)Inn(G).
Proof:
Define the map T :G/Z(G)→Inn(G) via
gZ(G)7→ .g
(We’ll use Gallian’s definition of inner auto-
1morphism: (x)=gxg .)g
We need to show that T is a well-defined func-
tion, that is one-to-one, onto, and preserves
the group operation.
8ToshowthatT iswell-defined,supposegZ(G)=
hZ(G). We’ll show that both cosets map to
the same inner automorphism.
1Now gZ(G) = hZ(G) implies h g ∈ Z(G).
1 1Then for all x∈G, h gx=xh g.
Bymultiplyingontheleftby handontheright
1 1 1by g , we have gxg =hxh .
Therefore = .g h
9One-to-one: reverse the argument. = g h
implies gZ(G)=hZ(G).
Onto: by definition of T. For any , the cosetg
gZ(G) is a preimage under T.
Group operation:
T(gZ(G)hZ(G))=T(ghZ(G))= .gh
T(gZ(G))T(hZ(G))= = .g h gh
1 1(Recall: (x)= (hxh )=(gh)x(gh) .)g gh
10